Practice MCQ Questions and Answer on Problems on Numbers
11.
A two-digit number is 7 times the sum of its two digits. The number that is formed by reversing its digits is 18 less than the original number. What is the number ?
(A) 42
(B) 52
(C) 62
(D) 72
Solution:
Let the ten's digit be x and the unit's digit be y Then, number = 10x + y $$\eqalign{ & \therefore 10x + y = 7\left( {x + y} \right) \cr & \Leftrightarrow 3x = 6y \cr & \Leftrightarrow x = 2y \cr} $$ Number formed by reversing the digits = 10y + x $$\eqalign{ & \therefore \left( {10x + y} \right) - \left( {10y + x} \right) = 18 \cr & \Leftrightarrow 9x - 9y = 18 \cr & \Leftrightarrow x - y = 2 \cr & \Leftrightarrow 2y - y = 2 \cr & \Leftrightarrow y = 2 \cr & {\text{So, }}x = 2y = 4 \cr} $$ Hence, ∴ Required number = 10x + y = 40 + 2 = 42
12.
The sum of four numbers is 64. If you add 3 to the first number, 3 is subtracted from the second number, the third is multiplied by 3 and the fourth is divided by 3, then all the results are equal. What is the difference between the largest and the smallest of the original numbers ?
(A) 21
(B) 27
(C) 32
(D) Cannot be determined
Solution:
Let the four numbers be , A, B, C and D Let A + 3 = B - 3 = 3C = $$\frac{D}{3}$$ = x Then, A = x - 3 B = x + 3 C = $$\frac{x}{3}$$ D = 3x $$\eqalign{ & \Leftrightarrow A + B + C + D = 64 \cr & \Leftrightarrow \left( {x - 3} \right) + \left( {x + 3} \right) + \frac{x}{3} + 3x = 64 \cr & \Leftrightarrow 5x + \frac{x}{3} = 64 \cr & \Leftrightarrow 16x = 192 \cr & \Leftrightarrow x = 12 \cr} $$ Thus, the numbers are 9, 15, 4 and 36 ∴ Required difference : = (36 - 4) = 32
13.
The sum of seven consecutive numbers is 175. What is the difference between twice the largest number and thrice the smallest number ?
The difference between a two-digit number and the number obtained by interchanging the positions of its digits is 36. What is the difference between the two digits of the number ?
(A) 3
(B) 4
(C) 9
(D) Cannot be determined
Solution:
Let the ten's digit be x and unit's digit be y Then, $$\eqalign{ & \Leftrightarrow \left( {10x + y} \right) - \left( {10y + x} \right) = 36 \cr & \Leftrightarrow 9\left( {x - y} \right) = 36 \cr & \Leftrightarrow x - y = 4 \cr} $$
15.
In a two-digit number, if it is known that its unit's digits exceeds its ten's digit by 2 and that the product of the given number and the sum of its digits is equal to 144, then the number is :
The digit in the unit's place of a numbers is equal to the digit in the ten's place of half of that number and the digit in the ten's place of that number is less than the digit in unit's place of half of the number by 1. If the sum of the digits of the number is 7, then what is the number ?
(A) 34
(B) 52
(C) 162
(D) Data inadequate
Solution:
Let the ten's digit be x and unit's digit be y Then, $$\eqalign{ & \Leftrightarrow \frac{{10x + y}}{2} = 10y + \left( {x + 1} \right) \cr & \Leftrightarrow 10x + y = 20y + 2x + 2 \cr & \Leftrightarrow 8x - 19y = 2.....(i) \cr & {\text{And}} \Leftrightarrow x + y = 7.....(ii) \cr} $$ Solving (i) and (ii), we get : x = 5, y = 2 Hence, required number = 52
17.
A two-digit number is such that the product of the digits is 8. When 18 is added to the number, then the digits are reversed. The number is :
(A) 18
(B) 24
(C) 42
(D) 81
Solution:
Let the ten's and unit's digits be $$x$$ and $$\frac{8}{x}$$ respectively Then, $$\eqalign{ & \Leftrightarrow \left( {10x + \frac{8}{x}} \right) + 18 = 10 \times \frac{8}{x} + x \cr & \Leftrightarrow 10{x^2} + 8 + 18x = 80 + {x^2} \cr & \Leftrightarrow 9{x^2} + 18x - 72 = 0 \cr & \Leftrightarrow {x^2} + 2x - 8 = 0 \cr & \Leftrightarrow \left( {x + 4} \right)\left( {x - 2} \right) = 0 \cr & \Leftrightarrow x = 2 \cr} $$ So, ten's digit = 2 and unit's digit = 4 Hence, required number = 24
18.
A certain number of two digits is three times the sum of its digits and if 45 be added to it, the digits are reversed. The number is ?
(A) 23
(B) 27
(C) 32
(D) 72
Solution:
Let the ten's digit be x and unit's digit be y Then, $$\eqalign{ & \Rightarrow 10x + y = 3\left( {x + y} \right) \cr & \Rightarrow 7x - 2y = 0.....(i) \cr & 10x + y + 45 = 10y + x \cr & \Rightarrow y - x = 5.....(ii) \cr} $$ Solving (i) and (ii), we get : x = 2 and y = 7 ∴ Required number = 27
19.
A number whose fifth part increase by 4 is equal to its fourth part diminished by 10, is :
(A) 240
(B) 260
(C) 270
(D) 280
Solution:
Let the number be x Then, $$\eqalign{ & \Leftrightarrow \frac{x}{5} + 4 = \frac{x}{4} - 10 \cr & \Leftrightarrow \frac{x}{4} - \frac{x}{5} = 14 \cr & \Leftrightarrow \frac{x}{{20}} = 14 \cr & \Leftrightarrow x = 14 \times 20 \cr & \Leftrightarrow x = 280 \cr} $$
20.
The sum of the digits of a two-digit number is 15 and the difference between the digits is 3. What is the two-digit number?
(A) 69
(B) 78
(C) 96
(D) Cannot be determined
Solution:
Let the ten's digit be x and unit's digit be y Then, x + y = 15 and x - y = 3 or y - x = 3 Solving x + y = 15 and x - y = 3, we get: x = 9, y = 6 Solving x + y = 15 and y - x = 3, we get: x = 6, y = 9 So, the number is either 96 or 69 Hence, the number cannot be determined.